Question

wave length of two notes in air is (90/175) m and (90/173) m , respectively . Each of these notes produces 4 beats/s with a third note of a fixed frequency . calculate the velocity of sound in air

Answer 1

Hello pari : -) 

  let lambda 1= > T1 and lambda 2 = > T2 

 given T1 =( 90/175 ) m         and T2 = > (90/173) m 

if F1 and F2 are the corrosponding frequencies and v is the velocity of sound in air , we have 

v = > F1T1        and v = > F2T2 

F1 = > v / T1      and F2 = > v/T2

since T1 < T2 we must have F1 > F2 

if F is the frequency of the third note , then 

F1 - f = > 4     and f - F2 = > 4 

we get F1- F2  = > 8 

v/T1    -    v/T2 = > 8 

v{ 175/90 -  173/90 }  => 8 

v ×2/90 = > 8 we get v = > 360 m /s   is our required answer 

Hopes this will help you pari !

@ engineer gopal khandelwal 

Answer 2

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Answer:

$\quad\displaystyle\sf{\lambda_1=\dfrac{90}{175} \: \: and \: \: \lambda_2=\dfrac{90}{173} }$

$\quad\displaystyle\sf{v=f_1 \lambda_1 \: and \: v=f_2 \lambda_2}$

•If f1 and f 2 are corresponding frequency and v is the velocity of the surrounding to air we have,

$\implies\displaystyle\sf{f_1=\dfrac{v}{\lambda_1} \: \: and \: \: f_2=\dfrac{v}{\lambda_2}}$

•since$\sf{\lambda_1 < \lambda_2}$, we must have $\sf{f_1 > f_2}$

$\\ \implies\displaystyle\sf{ f_1-f_2=4 ...(1) \: \: and \: \: f_1-f_2=4 ......(2)}$

$\implies\displaystyle\sf{f_1-f_2=8}$

$\\ \implies\displaystyle\sf{\dfrac{v}{\lambda_1} - \dfrac{v}{\lambda_2}=8}$

$\implies\displaystyle\sf{ v \bigg(\dfrac{175}{90} -\dfrac{173}{90}\bigg) }$

$\implies\displaystyle\sf{v=360 m/s}$

$\boxed{\underline{\displaystyle{\bf{velocity=360 m/s}}}}$